I'm developing a device with changeable AAA batteries running at 3V. Since the end user can change the batteries there is a chance they may be put in the wrong way around. The circuit basically consists of an ATTiny85 and some LEDs. Do I need to add a reverse polarity protection circuit to my design or should it be fine without?

If it needs one, what is the simplest way to achieve this? Just a single diode between +3V and Vcc on the MCU should do, right?

  • 2
    \$\begingroup\$ The ATtiny85 is only speced down to about 2.7V. For operation on 2 AAAs (especially rechargables) you'll probably want to use the ATtiny85V instead. \$\endgroup\$ Commented Jul 1, 2015 at 17:17
  • \$\begingroup\$ I didn't know there was a low voltage version of the ATTiny85. Thank you, this helps me! \$\endgroup\$ Commented Jul 1, 2015 at 17:28
  • \$\begingroup\$ There's a nice trick using an NMOS instead of a diode for reverse protection. This way you don't get the diode drop, which you really don't want in a battery device. \$\endgroup\$
    – avl_sweden
    Commented Jul 1, 2015 at 17:30
  • \$\begingroup\$ I heard it was with a PMOS. \$\endgroup\$ Commented Jul 1, 2015 at 17:41
  • 1
    \$\begingroup\$ Another approach is physical protection. Some battery compartments have ridges around the positive terminal. With a correctly inserted battery, the bump on the positive end of the battery will go between the ridges and make contact. However, with a backwards battery, the flat negative end will not make contact. If you have this arrangement, you have little need to electronic protection. \$\endgroup\$
    – DoxyLover
    Commented Jul 1, 2015 at 18:01

2 Answers 2


A better solution for reverse polarity protection using a P-channel MOSFET:


simulate this circuit – Schematic created using CircuitLab

This circuit has virtually no voltage drop across the MOSFET, where-as a diode will have one (even a schottky diode might have a ~0.25V drop).

Note that this circuit requires that \$V_1 > V_{th}\$, which in this case would be ~2V. Also with 2AAA batteries you might have a hard time driving some LED's, especially as the batteries drain.

If you prefer, here's an equivalent circuit which uses an N-channel MOSFET. Similarly, it requires \$V_1 > V_{th}\$


simulate this circuit

For a more complete explanation on why these circuits work, I would recommend watching this (it only covers the PMOS circuit, but a similar concept applies to the NMOS circuit).

Another related question: Is a Schottky diode appropriate for reverse polarity protection?


enter image description here

Operating Voltage For ATtiny85 is 2.7 to 5.5v. You are trying to run your system with just two AA batterries i.e aprroximately 3.2v (Battery voltage is a bit higher than rated not just 1.5X2=3V). So when you put a silicon diode in series then you would loose approx. 0.7V. That means you would only have (3.2-0.7=2.5V) which is lower than what is specified on Datasheet. Moreover, if you are drawing 100ma of current then it dissipates 0.7x100ma=70mW of power which is too much for a battery powered system. A schottkey diode has a voltage drop of about 0.25V. It may do the job.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.